"""
给你一个字符串 s 和一个字符规律 p，请你来实现一个支持 '.' 和 '*' 的正则表达式匹配。
'.' 匹配任意单个字符
'*' 匹配零个或多个前面的那一个元素
所谓匹配，是要涵盖 整个 字符串 s的，而不是部分字符串。

Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:
'.' Matches any single character.​​​​
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).

"""

"""
本题有动态规划、状态机、递归三种高阶解法，本人还不太熟悉，还在研究中。。。
"""
class Solution:
    def isMatch(self, s: str, p: str) -> bool:

        m, n = len(s), len(p)
        dp = [[False] * (n+1) for _ in range(m+1)]
        
        # 初始化
        dp[0][0] = True
        for j in range(1, n+1):
            if p[j-1] == '*':
                dp[0][j] = dp[0][j-2]

        # 状态更新
        for i in range(1, m+1):
            for j in range(1, n+1):
                if s[i-1] == p[j-1] or p[j-1] == '.':
                    dp[i][j] = dp[i-1][j-1]
                elif p[j-1] == '*':    
                    if s[i-1] != p[j-2] and p[j-2] != '.':
                        dp[i][j] = dp[i][j-2]
                    else:
                        dp[i][j] = dp[i][j-2] | dp[i-1][j]
        
        return dp[m][n]